Mat*_*oli 11 php doctrine zend-framework doctrine-orm
我正在整合Zend Framework和Doctrine 2.
问题是,在我的控制器和视图中,需要访问模型.我可以通过EntityManager的单个实例完成所有这些操作.
我在哪里存储这个实例?
Zend_Registry?这就是它现在的位置,它可以从任何地方访问,但不是很实用:$em = Zend_Registry::get('EntityManager');$this->em,我喜欢这个$em = My\EntityManager\Factory::getInstance();.封装很好,但很长时间打字......EntityManager一个单身了吗?- >(更新)不是不是Cob*_*bby 12
我不建议直接在您的控制器和视图中使用EntityManager.相反,使用服务层并注入EntityManager.
我有两个自定义操作助手,一个用于检索存储库,另一个用于服务.每个操作都包含对EntityManager的引用,并在将其返回给Controller之前将其注入.
不是我的实际代码,而是这样的(未经测试):
我的/控制器/动作/助手/ Service.php
<?php
namespace My\Controller\Action\Helper;
use Doctrine\ORM\EntityManager;
class Service extends \Zend_Controller_Action_Helper_Abstract
{
private $em;
public function __construct(EntityManager $em)
{
$this->em = $em;
}
public function direct($serviceClass)
{
return new $serviceClass($this->em);
}
}
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您可以编写类似的Action Helper来检索存储库.
然后,在您的引导程序中注册帮助程序(我们也可以访问EntityManager):
<?php
use Zend_Controller_Action_HelperBroker as HelperBroker,
My\Controller\Action\Helper\Service;
class Bootstrap extends Zend_Application_Bootstrap_Bootstrap
{
public function _initActionHelpers()
{
$this->bootstrap('doctrine');
$em = $this->getResource('doctrine');
HelperBroker::addHelper(new Service($em));
}
}
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现在写一个简单的服务.
我/域/博客/服务/ PostService.php
<?php
namespace My\Domain\Blog\Service;
use Doctrine\ORM\EntityManager,
My\Domain\Blog\Entity\Post;
class PostService implements \My\Domain\Service
{
private $em;
public function __construct(EntityManager $em)
{
$this->em = $em;
}
public function createPost($data)
{
$post = new Post();
$post->setTitle($data['title']);
$post->setContent($data['content']);
$this->em->persist($post);
$this->em->flush(); // flush now so we can get Post ID
return $post;
}
}
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并在控制器动作中将它们组合在一起:
<?php
class Blog_PostController extends Zend_Controller_Action
{
private $postService;
public function init()
{
$this->postService = $this->_helper->Service('My\Domain\Blog\PostService');
}
public function createPostAction()
{
// normally we'd get data from the actual request
$data = array(
"title" => "StackOverflow is great!",
"content" => "Imagine where I'd be without SO :)"
);
// and then validate it too!!
$post = $this->postService->createPost($data);
echo $post->getId(); // Blog post should be persisted
}
}
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